Question

A 20.0 gram aluminum ball (specific heat 0.90 J g‐1 °C‐1) at 80 °C was placed in 50 grams of water initially at 22.9 °C. What is the final temperature of the water and aluminum assuming no heat was lost during the energy exchange.

Answer 1

Answer on question #54862 - Chemistry - General chemistry

Question:

A 20.0 gram aluminum ball (specific heat 0.90 J g⁻¹ °C⁻¹) at 80 °C was placed in 50 grams of water initially at 22.9 °C. What is the final temperature of the water and aluminum assuming no heat was lost during the energy exchange.

Solution:

The energy amount going out of the warm water is equal to the energy amount going into the cool water. This means:

$$q_{\text{lost}} = q_{\text{gain}}$$ $$q = m \times C \times \Delta T$$

by substitution:

$$m_1 \times C_1(T_1 - x) = m_2 \times C_2(x - T_2)$$ $$20 \times 0.9 \times (80 - x) = 50 \times 1(x - 22.9)$$ $$1440 - 18x = 50x - 1145$$ $$68x = 2585$$ $$x = 38.01$$

Answer:

the final temperature of water and aluminum is 38.01 °C

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